Question

What is the equation of the tangent line of `f(x) = (x^3 - 3x + 1)(x + 2)` at the given point of (1, -3) ?

Answer 1

`y=-x-2`

Explanation:

Find the derivative of the function.

Begin by using the Product Rule.

`f'(x)=uv'+u'v`

`u=x^3-3x+1`
`u'=3x^2-3`

`v=x+2`
`v'=1`

`f'(x)=(x^3-3x+1)(1)+(3x^2-3)(x+2)`

`f'(x)=(x^3-3x+1)+(3x^2-3)(x+2)`

Substitute in the value of `x` to get a numerical value for the slope, `m`.

`f'(1)=((1)^3-3(1)+1)+(3(1)^2-3)((1)+2)`

Simplify

`f'(1)=(1-3+1)+(3-3)(1+2)`

`f'(1)=-1+(0)(3)=-1+0=-1`

The slope, `m`, is `-1`.

Substitute in the point `(1,-3)` and the slope, `m`, `-1`, into the slope intercept formula, `y=mx+b`.

`-3=-1(1)+b`

Simplify

`-3=-1+b`

Add `1` to both sides of the equation to isolate `b`.

`-2=b`

Write the equation of the tangent line using the above information.

`y=-1x-2`

`or`

`y=-x-2`

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